Question

1. Find the solution using distributivity.

{ 7/5 x (-3/10) } + { 7/6 x 9/10 }

2. Simplify 2/5 + 3/7 + (-6/5) + (13/7).

3. Write the additive inverse of the following. (i) 5/8 (ii) -9/7.

4. Find 2/5 x (-1/9) + 23/180 – 1/9 x 3/4.

5. Find the product of 2/5 and reciprocal of 25/4.​

Answer 1

/* First question must be like this */

$\red{ 1. \frac{7}{5} \times \Big(\frac{-3}{10}\Big) + \frac{7}{5} \times \frac{9}{10}} \\= \frac{7}{5} \Big ( \frac{-3}{10} + \frac{9}{10} \Big) \\= \frac{7}{5} \Big ( \frac{-3+9}{10} \Big)$

/* ________________

By Distributive Property :

$\pink { A\times B + A \times C = A( B+C) }$

____________________*/

$= \frac{7}{5} \times \frac{6}{10}$

$= \frac{7}{5} \times \frac{3}{5}$

$= \frac{21}{25}$

Therefore.,

$\red{ \frac{7}{5} \times \Big(\frac{-3}{10}\Big) + \frac{7}{5} \times \frac{9}{10}}\green {= \frac{21}{25}}$

$\red{ 2. \frac{2}{5} + \frac{3}{7 }+ \big(\frac{-6}{5}\big) + \frac{13}{7} }$

/* Rearranging the terms, we get */

$= \frac{2}{5} - \frac{6}{5 }+ \frac{3}{7} + \frac{13}{7}$

$= \frac{2-6}{5} + \frac{3+13}{7}$

$= \frac{-4}{5} + \frac{16}{7}$

$= \frac{ -28+80}{35} \\ = \frac{ 52}{35}$

Therefore.,

$\red{ \frac{2}{5} + \frac{3}{7 }+ \big(\frac{-6}{5}\big) + \frac{13}{7} }\green { = \frac{ 52}{35}}$

$\red{ 3.i) Additive \: inverse \: of \: \frac{5}{8 } } \\is \:\underline { \blue { \frac{-5}{8}}}$

$\red{ 3. ii)Additive \: inverse \: of \: \frac{-9}{7 } } \\is \:\underline { \blue { \frac{9}{7}}}$

$\red{4.\frac{2}{5} \times \Big(\frac{-1}{9}\Big) + \frac{23}{180} - \frac{1}{9} \times \frac{3}{4}}$

/* Rearranging the terms,we get */

$= \frac{2}{5} \times \big(\frac{-1}{9}\big) + \frac{- 1}{9}\times \frac{3}{4}+ \frac{23}{180}$

$= \frac{-1}{9} \Big ( \frac{2}{5} + \frac{3}{4} \Big) + \frac{23}{180}$

$= \frac{-1}{9} \times \frac{8+15}{20} + \frac{23}{180}$

$= \frac{-1}{9} \times \frac{23}{20} + \frac{23}{180}$

$= (-1)\times \frac{23}{9\times 20} + \frac{23}{180}$

$= \frac{-23}{180} + \frac{23}{180}$

$= 0$

Therefore.,

$\red{\frac{2}{5} \times \Big(\frac{-1}{9}\Big) + \frac{23}{180} - \frac{1}{9} \times \frac{3}{4}}\green {=0}$

$\red{ 5 . The\: product\: of \:\frac{2}{5}\: and}\\\red{ reciprocal \:of \: \frac{25}{4} }\\= \frac{2}{5} \times \blue { \frac{4}{25}} \\= \frac{2 \times 4 }{5 \times 25 } \\= \frac{8}{125}$

Therefore.,

$\red{ The\: product\: of \:\frac{2}{5}\: and}\\\red{ reciprocal \:of \: \frac{25}{4} } \\\green {= \frac{8}{125}}$

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